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Q.No. Question Choose the correct option for each of the following. (1) If f(x) = x⁴ / (x-1) + x² / (x-1) then f(x) is equal to (a) 0 (b) x⁴ (c) 2x² (d) 1 (II) If tan 5x = kx, then the value of k is (a) 0 (b) 5 (c) 5x (d

Concept: Continuity of a function

Question

Q.No. Question
Choose the correct option for each of the following.
(1) If f(x) = x⁴ / (x-1) + x² / (x-1) then f(x) is equal to
(a) 0
(b) x⁴
(c) 2x²
(d) 1
(II) If tan 5x = kx, then the value of k is
(a) 0
(b) 5
(c) 5x
(d) no
(III) The domain of the function f(x) = √(x-2) - √(5-x) is
(a) [2,5]
(b) (2,5)
(c) (-∞, 2) U (5,∞)
(d) [5,∞) only
(IV) d/dx (sin²x) is equal to
(a) cos²x
(b) 3x² cos²x
(c) -3x² cos²x
(d) 3x cos²x
(V) If y = tan⁻¹ x, then the value of dy/dx at x=0 is
(a) 1
(b) 0
(c) -1
(d) 1
(VI) The function f(x) = |x-3| is continuous
(a) for all real numbers x
(b) only at x=-3
(c) only at x=0
(d) nowhere
2. Evaluate: lim (x²-9) / (x²-4)
OR
For what value of k is the following function continuous at x=5?
f(x) = { (x²-25)/(x-5) if x≠5
{ k if x=5
3. Differentiate cos(sin(√x)) with respect to x
OR
Differentiate y = log (x + √(x²-a²)) with respect to x.
4. If xʸ = (x+y)ˣ, prove that dy/dx = y(y-x log y) / x(x-y log x)
OR
Find the derivative of tan x with respect to x from the first principle.

Try it yourself first

Hint: Recall the definition of continuity at a point. What must be true about the limit and the function's value at x=5x=5?

Step-by-step solution

  1. 1

    Given function

    Function f(x)f(x) is defined piecewise.
    f(x)={x2−25x−5if x≠5kif x=5f(x) = \begin{cases} \frac{x^2-25}{x-5} & \text{if } x \neq 5 \\ k & \text{if } x=5 \end{cases}
  2. 2

    Continuity condition

    For continuity at x=ax=a, lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).
    lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)
  3. 3

    Apply condition at $x=5$

    We need lim⁡x→5f(x)=f(5)\lim_{x \to 5} f(x) = f(5).
    lim⁡x→5f(x)=f(5)\lim_{x \to 5} f(x) = f(5)
  4. 4

    Evaluate $f(5)$

    From the definition, f(5)=kf(5) = k.
    f(5)=kf(5) = k
  5. 5

    Evaluate $\lim_{x \to 5} f(x)$

    For x≠5x \neq 5, f(x)=x2−25x−5f(x) = \frac{x^2-25}{x-5}.
    lim⁡x→5x2−25x−5\lim_{x \to 5} \frac{x^2-25}{x-5}
  6. 6

    Simplify the limit

    Factorize the numerator and cancel (x−5)(x-5).
    lim⁡x→5(x−5)(x+5)x−5=lim⁡x→5(x+5)\lim_{x \to 5} \frac{(x-5)(x+5)}{x-5} = \lim_{x \to 5} (x+5)
  7. 7

    Calculate the limit

    Substitute x=5x=5 into the simplified expression.
    lim⁡x→5(x+5)=5+5=10\lim_{x \to 5} (x+5) = 5+5 = 10
  8. 8

    Equate for continuity

    For continuity, k=10k = 10.
    k=10k = 10

Final answer

10

Remember: For continuity, function value must equal the limit at that point.

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For a function f(x)f(x) to be continuous at x=ax=a, which of the following conditions must be met?

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